Difference between revisions of "2021 Fall AMC 12B Problems/Problem 14"
(Created page with "==Problem== Suppose that <math>P(z), Q(z)</math>, and <math>R(z)</math> are polynomials with real coefficients, having degrees <math>2</math>, <math>3</math>, and <math>6</mat...") |
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==Solution== | ==Solution== | ||
− | The answer | + | The answer cannot be <math>0</math>, as every nonconstant polynomial has at least <math>1</math> distinct complex root (fundamental theorem of algebra); the polynomial <math>P(z)\cdot Q(z)</math> has degree <math>2 + 3 = 5</math>, so the polynomial <math>R(z) - P(z)\cdot Q(z)</math> has degree <math>6</math> and is thus nonconstant. It now suffices to illustrate an example for which <math>N = 1</math>. |
Take <math>P(z)=z^2+1,Q(z)=z^3+2,</math> and <math>R(z)=(z+1)^6 + \left(z^2+1\right)\left(z^3+2\right).</math> | Take <math>P(z)=z^2+1,Q(z)=z^3+2,</math> and <math>R(z)=(z+1)^6 + \left(z^2+1\right)\left(z^3+2\right).</math> | ||
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− | ~kingofpineapplz | + | ~kingofpineapplz and kgator |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2021 Fall|ab=B|num-a=13|num-b=11}} | {{AMC12 box|year=2021 Fall|ab=B|num-a=13|num-b=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:09, 16 January 2022
Problem
Suppose that , and are polynomials with real coefficients, having degrees , , and , respectively, and constant terms , , and , respectively. Let be the number of distinct complex numbers that satisfy the equation . What is the minimum possible value of ?
Solution
The answer cannot be , as every nonconstant polynomial has at least distinct complex root (fundamental theorem of algebra); the polynomial has degree , so the polynomial has degree and is thus nonconstant. It now suffices to illustrate an example for which . Take and
has degree 6 and constant term , so it satisfies the conditions. We need to find the solutions to or Clearly, there is one distinct complex root, , so our answer is
~kingofpineapplz and kgator
See Also
2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.